Hello!
The graph is attached.
To graph a parabola we need to know the following:
- If the parabola is open upwards or downwards
- They axis intercepts (if they exist)
- The vertex position (point)
We are given the function:
[tex]f(t)=-16t^{2}+64t+4[/tex]
Where,
[tex]a=-16\\b=64\\c=4[/tex]
For this case, the coefficient of the quadratic term (a) is negative, it means that the parabola opens downwards.
Finding the axis interception points:
Making the function equal to 0, we can find the x-axis (t) intercepts, but since the equation is a function of the time, we will only consider the positive values, so:
[tex]f(t)=-16t^{2}+64t+4\\0=-16t^{2}+64t+4\\-16t^{2}+64t+4=0[/tex]
Using the quadratic equation:
[tex]\frac{-b+-\sqrt{b^{2}-4ac } }{2a}=\frac{-64+-\sqrt{64^{2}-4*-16*4} }{2*-16}\\\\\frac{-64+-\sqrt{64^{2}-4*-16*4} }{2*-16}=\frac{-64+-\sqrt{4096+256} }{-32}\\\\\frac{-64+-\sqrt{4096+256} }{-32}=\frac{-64+-(65.96) }{-32}\\\\t1=\frac{-64+(65.96) }{-32}=-0.06\\\\t2=\frac{-64-(65.96) }{-32}=4.0615[/tex]
So, at t=4.0615 the height of the softball will be 0.
Since we will work only with positive values of "x", since we are working with a function of time:
Let's start from "t" equals to 0 to "t" equals to 4.0615.
So, evaluating we have:
[tex]f(0)=-16(0)^{2}+64(0)+4=4\\\\f(1)=-16(1)^{2}+64(1)+4=52\\\\f(2)=-16(2)^{2}+64(2)+4=68\\\\f(3)=-16(3)^{2}+64(3)+4=52\\\\f(4.061)=-16(4.0615)^{2}+64(4.0615)+4=0.0034=0[/tex]
Finally, we can conclude that:
- The softball reach its maximum height at t equals to 2. (68 feet)
- The softball hits the ground at t equals to 4.0615 (0 feet)
- At t equals to 0, the height of the softball is equal to 4 feet.
See the attached image for the graphic.
Have a nice day!
